Average Error: 0.5 → 0.5
Time: 13.7s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := 2 \cdot \left(n \cdot \pi\right)\\ \sqrt{\frac{1}{k}} \cdot \left({t_0}^{\left(k \cdot -0.5\right)} \cdot \sqrt{t_0}\right) \end{array} \]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\begin{array}{l}
t_0 := 2 \cdot \left(n \cdot \pi\right)\\
\sqrt{\frac{1}{k}} \cdot \left({t_0}^{\left(k \cdot -0.5\right)} \cdot \sqrt{t_0}\right)
\end{array}

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* 2.0 (* n PI))))
   (* (sqrt (/ 1.0 k)) (* (pow t_0 (* k -0.5)) (sqrt t_0)))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
↓
double code(double k, double n) {
	double t_0 = 2.0 * (n * ((double) M_PI));
	return sqrt(1.0 / k) * (pow(t_0, (k * -0.5)) * sqrt(t_0));
}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.5

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

  2. Simplified0.5

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}} \]

  3. Taylor expanded in n around -inf 64.0

    \[\leadsto \color{blue}{\sqrt{\frac{1}{k}} \cdot e^{\left(\log \left(-2 \cdot \pi\right) - \log \left(\frac{-1}{n}\right)\right) \cdot \left(0.5 - 0.5 \cdot k\right)}} \]

  4. Simplified0.5

    \[\leadsto \color{blue}{\sqrt{\frac{1}{k}} \cdot {\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}} \]

  5. Applied fma-udef_binary640.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot {\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{\color{blue}{\left(k \cdot -0.5 + 0.5\right)}} \]

  6. Applied unpow-prod-up_binary640.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\left({\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{\left(k \cdot -0.5\right)} \cdot {\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{0.5}\right)} \]

  7. Simplified0.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \left(\color{blue}{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)}} \cdot {\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{0.5}\right) \]

  8. Simplified0.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \left({\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)} \cdot \color{blue}{\sqrt{2 \cdot \left(n \cdot \pi\right)}}\right) \]

  9. Final simplification0.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \left({\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}\right) \]

Reproduce

herbie shell --seed 2021313 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))